QuestionMay 3, 2025

A skier starts from rest at the top of a hill that is inclined at 10.6^circ with respect to the horizontal. The hillside is 250 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest? square lm

A skier starts from rest at the top of a hill that is inclined at 10.6^circ with respect to the horizontal. The hillside is 250 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest? square lm
A skier starts from rest at the top of a hill that is inclined at 10.6^circ with respect to the horizontal. The hillside is 250 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest? square lm

Solution
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Answer

70.9 \text{ m} Explanation 1. Calculate gravitational force component along the incline The gravitational force component along the incline is mg \sin(\theta). 2. Calculate normal force on the incline Normal force is mg \cos(\theta). 3. Calculate frictional force on the incline Frictional force is \mu mg \cos(\theta), where \mu = 0.0750. 4. Calculate net force on the incline Net force is mg \sin(\theta) - \mu mg \cos(\theta). 5. Calculate acceleration on the incline Using F = ma, acceleration a = g (\sin(\theta) - \mu \cos(\theta)). 6. Calculate final velocity at the bottom of the hill Use v^2 = u^2 + 2as, with u = 0, s = 250 \text{ m}, a = g (\sin(\theta) - \mu \cos(\theta)). 7. Calculate deceleration on the horizontal surface Deceleration a_h = \mu g. 8. Calculate distance traveled on the horizontal surface Use v^2 = u^2 + 2a_h s_h, with v from Step 6, u = 0, and solve for s_h.

Explanation

1. Calculate gravitational force component along the incline<br /> The gravitational force component along the incline is $mg \sin(\theta)$. <br /><br />2. Calculate normal force on the incline<br /> Normal force is $mg \cos(\theta)$.<br /><br />3. Calculate frictional force on the incline<br /> Frictional force is $\mu mg \cos(\theta)$, where $\mu = 0.0750$.<br /><br />4. Calculate net force on the incline<br /> Net force is $mg \sin(\theta) - \mu mg \cos(\theta)$.<br /><br />5. Calculate acceleration on the incline<br /> Using $F = ma$, acceleration $a = g (\sin(\theta) - \mu \cos(\theta))$.<br /><br />6. Calculate final velocity at the bottom of the hill<br /> Use $v^2 = u^2 + 2as$, with $u = 0$, $s = 250 \text{ m}$, $a = g (\sin(\theta) - \mu \cos(\theta))$.<br /><br />7. Calculate deceleration on the horizontal surface<br /> Deceleration $a_h = \mu g$.<br /><br />8. Calculate distance traveled on the horizontal surface<br /> Use $v^2 = u^2 + 2a_h s_h$, with $v$ from Step 6, $u = 0$, and solve for $s_h$.
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